In the previous section we looked at building a binary search tree. As
we learned, the performance of the binary search tree can degrade to
`get`

and `put`

when the tree
becomes unbalanced. In this section we will look at a special kind of
binary search tree that automatically makes sure that the tree remains
balanced at all times. This tree is called an **AVL tree** and is named
for its inventors: G.M. Adelson-Velskii and E.M. Landis.

An AVL tree implements the Map abstract data type just like a regular
binary search tree, the only difference is in how the tree performs. To
implement our AVL tree we need to keep track of a **balance factor** for
each node in the tree. We do this by looking at the heights of the left
and right subtrees for each node. More formally, we define the balance
factor for a node as the difference between the height of the left
subtree and the height of the right subtree.

Using the definition for balance factor given above we say that a
subtree is left-heavy if the balance factor is greater than zero. If the
balance factor is less than zero then the subtree is right heavy. If the
balance factor is zero then the tree is perfectly in balance. For
purposes of implementing an AVL tree, and gaining the benefit of having
a balanced tree we will define a tree to be in balance if the balance
factor is -1, 0, or 1. Once the balance factor of a node in a tree is
outside this range we will need to have a procedure to bring the tree
back into balance. *Figure 1* shows an example of an unbalanced,
right-heavy tree and the balance factors of each node.

Before we proceed any further lets look at the result of enforcing this
new balance factor requirement. Our claim is that by ensuring that a
tree always has a balance factor of -1, 0, or 1 we can get better Big-O
performance of key operations. Let us start by thinking about how this
balance condition changes the worst-case tree. There are two
possibilities to consider, a left-heavy tree and a right heavy tree. If
we consider trees of heights 0, 1, 2, and 3, *Figure 2*
illustrates the most unbalanced left-heavy tree possible under the new
rules.

Looking at the total number of nodes in the tree we see that for a tree
of height 0 there is 1 node, for a tree of height 1 there is

This recurrence may look familiar to you because it is very similar to
the Fibonacci sequence. We can use this fact to derive a formula for the
height of an AVL tree given the number of nodes in the tree. Recall that
for the Fibonacci sequence the

An important mathematical result is that as the numbers of the Fibonacci
sequence get larger and larger the ratio of

By replacing the Fibonacci reference with its golden ratio approximation we get:

If we rearrange the terms, and take the base 2 log of both sides and
then solve for

This derivation shows us that at any time the height of our AVL tree is
equal to a constant(1.44) times the log of the height of the tree. This
is great news for searching our AVL tree because it limits the search to

Now that we have demonstrated that keeping an AVL tree in balance is going to be a big performance improvement, let us look at how we will augment the procedure to insert a new key into the tree. Since all new keys are inserted into the tree as leaf nodes and we know that the balance factor for a new leaf is zero, there are no new requirements for the node that was just inserted. But once the new leaf is added we must update the balance factor of its parent. How this new leaf affects the parent’s balance factor depends on whether the leaf node is a left child or a right child. If the new node is a right child the balance factor of the parent will be reduced by one. If the new node is a left child then the balance factor of the parent will be increased by one. This relation can be applied recursively to the grandparent of the new node, and possibly to every ancestor all the way up to the root of the tree. Since this is a recursive procedure let us examine the two base cases for updating balance factors:

- The recursive call has reached the root of the tree.
- The balance factor of the parent has been adjusted to zero. You should convince yourself that once a subtree has a balance factor of zero, then the balance of its ancestor nodes does not change.

We will implement the AVL tree as a subclass of `BinarySearchTree`

. To
begin, we will override the `_put`

method and write a new
`updateBalance`

helper method. These methods are shown in
*Listing 1*. You will notice that the definition for `_put`

is
exactly the same as in simple binary search trees except for the additions of
the calls to `updateBalance`

on lines 7 and 13.

**Listing 1**

```
def _put(self,key,val,currentNode):
if key < currentNode.key:
if currentNode.hasLeftChild():
self._put(key,val,currentNode.leftChild)
else:
currentNode.leftChild = TreeNode(key,val,parent=currentNode)
self.updateBalance(currentNode.leftChild)
else:
if currentNode.hasRightChild():
self._put(key,val,currentNode.rightChild)
else:
currentNode.rightChild = TreeNode(key,val,parent=currentNode)
self.updateBalance(currentNode.rightChild)
def updateBalance(self,node):
if node.balanceFactor > 1 or node.balanceFactor < -1:
self.rebalance(node)
return
if node.parent != None:
if node.isLeftChild():
node.parent.balanceFactor += 1
elif node.isRightChild():
node.parent.balanceFactor -= 1
if node.parent.balanceFactor != 0:
self.updateBalance(node.parent)
```

The new `updateBalance`

method is where most of the work is done. This
implements the recursive procedure we just described. The
`updateBalance`

method first checks to see if the current node is out
of balance enough to require rebalancing (line 16). If that
is the case then the rebalancing is done and no further updating to
parents is required. If the current node does not require rebalancing
then the balance factor of the parent is adjusted. If the balance factor
of the parent is non-zero then the algorithm continues to work its way
up the tree toward the root by recursively calling `updateBalance`

on
the parent.

When a rebalancing of the tree is necessary, how do we do it? Efficient
rebalancing is the key to making the AVL Tree work well without
sacrificing performance. In order to bring an AVL Tree back into balance
we will perform one or more **rotations** on the tree.

To understand what a rotation is let us look at a very simple example.
Consider the tree in the left half of *Figure 3*. This tree
is out of balance with a balance factor of -2. To bring this tree into
balance we will use a left rotation around the subtree rooted at node A.

To perform a left rotation we essentially do the following:

- Promote the right child (B) to be the root of the subtree.
- Move the old root (A) to be the left child of the new root.
- If new root (B) already had a left child then make it the right child of the new left child (A). Note: Since the new root (B) was the right child of A the right child of A is guaranteed to be empty at this point. This allows us to add a new node as the right child without any further consideration.

While this procedure is fairly easy in concept, the details of the code are a bit tricky since we need to move things around in just the right order so that all properties of a Binary Search Tree are preserved. Furthermore we need to make sure to update all of the parent pointers appropriately.

Lets look at a slightly more complicated tree to illustrate the right
rotation. The left side of *Figure 4* shows a tree that is
left-heavy and with a balance factor of 2 at the root. To perform a
right rotation we essentially do the following:

- Promote the left child (C) to be the root of the subtree.
- Move the old root (E) to be the right child of the new root.
- If the new root(C) already had a right child (D) then make it the left child of the new right child (E). Note: Since the new root (C) was the left child of E, the left child of E is guaranteed to be empty at this point. This allows us to add a new node as the left child without any further consideration.

Now that you have seen the rotations and have the basic idea of how a
rotation works let us look at the code. *Listing 2* shows the
code for both the right and the left rotations. In line 2
we create a temporary variable to keep track of the new root of the
subtree. As we said before the new root is the right child of the
previous root. Now that a reference to the right child has been stored
in this temporary variable we replace the right child of the old root
with the left child of the new.

The next step is to adjust the parent pointers of the two nodes. If
`newRoot`

has a left child then the new parent of the left child
becomes the old root. The parent of the new root is set to the parent of
the old root. If the old root was the root of the entire tree then we
must set the root of the tree to point to this new root. Otherwise, if
the old root is a left child then we change the parent of the left child
to point to the new root; otherwise we change the parent of the right
child to point to the new root. (lines 10-13).
Finally we set the parent of the old root to be the new root. This is a
lot of complicated bookkeeping, so we encourage you to trace through
this function while looking at *Figure 3*. The
`rotateRight`

method is symmetrical to `rotateLeft`

so we will leave
it to you to study the code for `rotateRight`

.

**Listing 2**

```
def rotateLeft(self,rotRoot):
newRoot = rotRoot.rightChild
rotRoot.rightChild = newRoot.leftChild
if newRoot.leftChild != None:
newRoot.leftChild.parent = rotRoot
newRoot.parent = rotRoot.parent
if rotRoot.isRoot():
self.root = newRoot
else:
if rotRoot.isLeftChild():
rotRoot.parent.leftChild = newRoot
else:
rotRoot.parent.rightChild = newRoot
newRoot.leftChild = rotRoot
rotRoot.parent = newRoot
rotRoot.balanceFactor = rotRoot.balanceFactor + 1 - min(newRoot.balanceFactor, 0)
newRoot.balanceFactor = newRoot.balanceFactor + 1 + max(rotRoot.balanceFactor, 0)
```

Finally, lines 16-17 require some explanation. In these two lines we update the balance factors of the old and the new root. Since all the other moves are moving entire subtrees around the balance factors of all other nodes are unaffected by the rotation. But how can we update the balance factors without completely recalculating the heights of the new subtrees? The following derivation should convince you that these lines are correct.

*Figure 5* shows a left rotation. B and D are the pivotal
nodes and A, C, E are their subtrees. Let

But we know that the old height of D can also be given by

and then subtract the two equations. The following steps
do the subtraction and use some algebra to simplify the equation for

Next we will move

But,

Now we have all of the parts in terms that we readily know. If we
remember that B is `rotRoot`

and D is `newRoot`

then we can see this
corresponds exactly to the statement on line 16, or:

```
rotRoot.balanceFactor = rotRoot.balanceFactor + 1 - min(0,newRoot.balanceFactor)
```

A similar derivation gives us the equation for the updated node D, as well as the balance factors after a right rotation. We leave these as exercises for you.

Now you might think that we are done. We know how to do our left and
right rotations, and we know when we should do a left or right rotation,
but take a look at *Figure 6*. Since node A has a balance
factor of -2 we should do a left rotation. But, what happens when we do
the left rotation around A?

*Figure 7* shows us that after the left rotation we are now
out of balance the other way. If we do a right rotation to correct the
situation we are right back where we started.

To correct this problem we must use the following set of rules:

- If a subtree needs a left rotation to bring it into balance, first check the balance factor of the right child. If the right child is left heavy then do a right rotation on right child, followed by the original left rotation.
- If a subtree needs a right rotation to bring it into balance, first check the balance factor of the left child. If the left child is right heavy then do a left rotation on the left child, followed by the original right rotation.

*Figure 8* shows how these rules solve the dilemma we
encountered in *Figure 6* and *Figure 7*. Starting
with a right rotation around node C puts the tree in a position where
the left rotation around A brings the entire subtree back into balance.

The code that implements these rules can be found in our `rebalance`

method, which is shown in *Listing 3*. Rule number 1 from
above is implemented by the `if`

statement starting on line 2.
Rule number 2 is implemented by the `elif`

statement starting on
line 8.

**Listing 3**

1 2 3 4 5 6 7 8 9 10 11 12 13 | ```
def rebalance(self,node):
if node.balanceFactor < 0:
if node.rightChild.balanceFactor > 0:
self.rotateRight(node.rightChild)
self.rotateLeft(node)
else:
self.rotateLeft(node)
elif node.balanceFactor > 0:
if node.leftChild.balanceFactor < 0:
self.rotateLeft(node.leftChild)
self.rotateRight(node)
else:
self.rotateRight(node)
``` |

The *discussion questions* provide you the opportunity to rebalance a tree
that requires a left rotation followed by a right. In addition the
discussion questions provide you with the opportunity to rebalance some
trees that are a little more complex than the tree in
*Figure 8*.

By keeping the tree in balance at all times, we can ensure that the
`get`

method will run in order `put`

method? Let us break this down
into the operations performed by `put`

. Since a new node is inserted
as a leaf, updating the balance factors of all the parents will require
a maximum of `put`

operation remains

At this point we have implemented a functional AVL-Tree, unless you need the ability to delete a node. We leave the deletion of the node and subsequent updating and rebalancing as an exercise for you.

Over the past two chapters we have looked at several data structures
that can be used to implement the map abstract data type. A binary
Search on a list, a hash table, a binary search tree, and a balanced
binary search tree. To conclude this section, let’s summarize the
performance of each data structure for the key operations defined by the
map ADT (see *Table 1*).

operation | Sorted List | Hash Table | Binary Search Tree | AVL Tree |
---|---|---|---|---|

put | ||||

get | ||||

in | ||||

del |